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8 tháng 11 2017

\(2\sqrt{3x-2}-2=11x+\sqrt{5x+6}-3\sqrt{\left(3x-2\right)\left(5x+6\right)}\)

ĐK: \(x\ge\dfrac{2}{3}\)

\(pt\Leftrightarrow2\sqrt{3x-2}-2-11x-\sqrt{5x+6}+3\sqrt{\left(3x-2\right)\left(5x+6\right)}=0\)

\(\Leftrightarrow2\sqrt{3x-2}-4-11x+22-\sqrt{5x+6}+4+3\sqrt{\left(3x-2\right)\left(5x+6\right)}-24=0\)

\(\Leftrightarrow2\dfrac{3x-2-16}{\sqrt{3x-2}+4}-11\left(x-2\right)-\dfrac{5x+6-16}{\sqrt{5x+6}+4}+\dfrac{9\left(3x-2\right)\left(5x+6\right)-576}{3\sqrt{\left(3x-2\right)\left(5x+6\right)}+24}=0\)

\(\Leftrightarrow\dfrac{6\left(x-2\right)}{\sqrt{3x-2}+4}-11\left(x-2\right)-\dfrac{5\left(x-2\right)}{\sqrt{5x+6}+4}+\dfrac{9\left(x-2\right)\left(15x+38\right)}{3\sqrt{\left(3x-2\right)\left(5x+6\right)}+24}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{6}{\sqrt{3x-2}+4}-11-\dfrac{5}{\sqrt{5x+6}+4}+\dfrac{9\left(15x+38\right)}{3\sqrt{\left(3x-2\right)\left(5x+6\right)}+24}\right)=0\)

\(\Rightarrow x-2=0\Rightarrow x=2\)

a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)

Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)

Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)

\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)

\(\Leftrightarrow b=a\)

Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)

\(\Leftrightarrow x^3-4x^2-6x+5=0\)

\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)

15 tháng 3 2016

ĐK : \(\begin{cases}x\ge\frac{-1}{3}\\y\le5\end{cases}\)

\(\sqrt{5x^2+3y+1}+1-4x=0\)

\(\Leftrightarrow\begin{cases}x\ge\frac{1}{4}\\5x^2+3y+1=16x^2-8x+1\left(1\right)\end{cases}\)

(1) \(\Leftrightarrow11x^2-8x-3y=0\left(2\right)\)

Đặt \(\begin{cases}\sqrt{3x+1}=a\left(a\ge0\right)\\\sqrt{5-y}=b\left(b\ge0\right)\end{cases}\) \(\Rightarrow\begin{cases}3x+2=a^2+1\\6-y=b^2+1\end{cases}\)

\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\\ \Leftrightarrow a^3-b^3+a-b=0\\ \Leftrightarrow\left(a-b\right)\left(a^2-ab+b^2+1\right)=0\\ \Leftrightarrow a-b=0\left(a^2-ab+b^2+1>0\right)\\\Leftrightarrow a=b\\ \)

\(\Rightarrow\sqrt{3x+1}=\sqrt{5-y}\\ \Leftrightarrow3x+1=5-y\\ \Leftrightarrow y=4-3x\left(3\right)\)

Từ (2) và (3)

 \(\Rightarrow11x^2-8x-3\left(4-3x\right)=0\\ \Leftrightarrow11x^2+x-12=0\\ \Leftrightarrow x=1\left(TM\right);x=\frac{-12}{11}\left(loại\right)\\ \Rightarrow y=1\left(TM\right)\)

Vậy S = \(\left\{\left(1;1\right)\right\}\)

14 tháng 3 2016

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